# C2 further differentiation

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#1
basically diff indices
6
----
5x^3 + 4x^3
now the 4x^3 bit is simply 12x squared but the six over 5x^3 is confusing me a bit, would it be just 6/5 times x^3 and then differentiate it. that seems too easy. any suggestions?
0
15 years ago
#2
rewrite it as (6/5)(x^-3) + 4(x^3)
0
#3
(Original post by El Stevo)
rewrite it as (6/5)(x^-3) + 4(x^3)
what i said then ok cool
0
15 years ago
#4
(Original post by Mr. Brightside)
basically diff indices
6
----
5x^3 + 4x^3
now the 4x^3 bit is simply 12x squared but the six over 5x^3 is confusing me a bit, would it be just 6/5 times x^3 and then differentiate it. that seems too easy. any suggestions?
d/dx {6/(5x^3) + 4x^3} = d/dx {(6/5)(x^-3) + 4x^3} = (6/5)(-3)(x^-4) + (4)(3)(x^2) = 12x^2 - 18/(5x^4) = (60x^6 - 18)/(5x^4)

Nima
0
15 years ago
#5
(Original post by Mr. Brightside)
what i said then ok cool
no... you said (6/5)*(x^3)... what i said has a very important difference... (6/5)*(x^-3)
0
15 years ago
#6
(Original post by El Stevo)
no... you said (6/5)*(x^3)... what i said has a very important difference... (6/5)*(x^-3)
his

6
----
5x^3

is 6/(5x^3) ?
0
15 years ago
#7
yes... but his
(Original post by Mr. Brightside)
would it be just 6/5 times x^3 and then differentiate it
is wrong...
0
15 years ago
#8
oh yeh ok then.
0
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